resnick 5ed vol2 (resuelto)

Instructor Solutions Manual
for
Physics
by
Halliday, Resnick, and Krane
Paul Stanley
Beloit College
Volume 2

A Note To The Instructor...
The solutions here are somewhat brief, as they are designed for the instructor, not for the student.
Check with the publishers before electronically posting any part of these solutions; website, ftp, or
server access must be restricted to yourstudents.
I have been somewhat casual about subscripts whenever it is obvious that a problem is one
dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.
Although this does not change the validity of the answer, it will sometimes obfuscate the approach
if viewed by a novice.
There are some traditional formula, such as
2
2
vx = v0x + 2ax x,

whichare not used in the text. The worked solutions use only material from the text, so there may
be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an
easier approach existed. But if it was not in the text, I did not use it here.
I also tried to avoid reinventing the wheel. There are some exercises and problems in the text
which build upon previous exercisesand problems. Instead of rederiving expressions, I simply refer
you to the previous solution.
I adopt a different approach for rounding of significant figures than previous authors; in particular, I usually round intermediate answers. As such, some of my answers will differ from those in
the back of the book.
Exercises and Problems which are enclosed in a box also appear in the Student’s SolutionManual
with considerably more detail and, when appropriate, include discussion on any physical implications
of the answer. These student solutions carefully discuss the steps required for solving problems, point
out the relevant equation numbers, or even specify where in the text additional information can be
found. When two almost equivalent methods of solution exist, often both are presented.You are
encouraged to refer students to the Student’s Solution Manual for these exercises and problems.
However, the material from the Student’s Solution Manual must not be copied.
Paul Stanley
Beloit College
stanley@clunet.edu

1

E25-1

The charge transferred is
Q = (2.5 × 104 C/s)(20 × 10−6 s) = 5.0 × 10−1 C.

E25-2 Use Eq. 25-4:
(8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C)= 1.40 m
(5.66 N)

r=
E25-3 Use Eq. 25-4:
F =

(8.99×109 N·m2 /C2 )(3.12×10−6 C)(1.48×10−6 C)
= 2.74 N.
(0.123 m)2

E25-4 (a) The forces are equal, so m1 a1 = m2 a2 , or
m2 = (6.31×10−7 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97×10−7 kg.
(b) Use Eq. 25-4:
q=

E25-5

(6.31×10−7 kg)(7.22 m/s2 )(3.20×10−3 m)2
= 7.20×10−11 C
(8.99×109 N·m2 /C2 )

(a) Use Eq. 25-4,
F =

1 q1 q2
1(21.3 µC)(21.3 µC)
=
= 1.77 N
2
−12 C2 /N · m2 )
4π 0 r12
4π(8.85×10
(1.52 m)2

(b) In part (a) we found F12 ; to solve part (b) we need to first find F13 . Since q3 = q2 and
r13 = r12 , we can immediately conclude that F13 = F12 .
We must assess the direction of the force of q3 on q1 ; it will be directed along the line which
connects the two charges, and will be directed away from q3 .The diagram below shows the directions.

F

θ

F 12
F

23

23

F net

F 12

From this diagram we want to find the magnitude of the net force on q1 . The cosine law is
appropriate here:
F net 2

F net

2
2
= F12 + F13 − 2F12 F13 cos θ,
2
= (1.77 N) + (1.77 N)2 − 2(1.77 N)(1.77 N) cos(120◦ ),
= 9.40 N2 ,
= 3.07 N.

2

E25-6 Originally F0 = CQ2 = 0.088 N, where C is aconstant. When sphere 3 touches 1 the
0
charge on both becomes Q0 /2. When sphere 3 the touches sphere 2 the charge on each becomes
(Q0 + Q0 /2)/2 = 3Q0 /4. The force between sphere 1 and 2 is then
F = C(Q0 /2)(3Q0 /4) = (3/8)CQ2 = (3/8)F0 = 0.033 N.
0
E25-7 The forces on q3 are F31 and F32 . These forces are given by the vector form of Coulomb’s
Law, Eq. 25-5,
F31

=

F32

=

1...